Section 6.3 Trigonometric Integrals
Subsection 6.3.1 Integrals of the form \(\int \sin^m(x) \cos^n(x) \, dx\)
Key Idea 6.3.2. Integrals Involving Powers of Sine and Cosine.
- If \(m\) is odd, then \(m=2k+1\) for some integer \(k\text{.}\) Rewrite\begin{align*} \sin^m(x) \amp = \sin^{2k+1}(x)\\ \amp = \sin^{2k}(x) \sin(x)\\ \amp = (\sin^2(x) )^k\sin(x)\\ \amp = (1-\cos^2(x) )^k\sin(x)\text{.} \end{align*}Then\begin{align*} \int \sin^m(x) \cos^n(x) \, dx \amp = \int (1-\cos^2(x) )^k\sin(x) \cos^n(x) \, dx\\ \amp = -\int (1-u^2)^ku^n\,du\text{,} \end{align*}where \(u = \cos(x)\) and \(du = -\sin(x) \, dx\text{.}\)
- If \(n\) is odd, then using substitutions similar to that outlined above (replacing all of the even powers of \(cosine\) using a Pythagorean identity) we have:\begin{equation*} \int \sin^m(x) \cos^n(x) \, dx = \int u^m(1-u^2)^k\,du\text{,} \end{equation*}where \(u = \sin(x)\) and \(du = \cos(x) \, dx\text{.}\)
- If both \(m\) and \(n\) are even, use the power-reducing identities:\begin{equation*} \cos^2(x) = \frac{1+\cos(2x)}{2} \text{ and } \sin^2(x) = \frac{1-\cos(2x)}2 \end{equation*}to reduce the degree of the integrand. Expand the result and apply the principles of this Key Idea again.
Example 6.3.3. Integrating powers of sine and cosine.
Solution 1.
Solution 2. Video solution
Example 6.3.4. Integrating powers of sine and cosine.
Solution.
Example 6.3.6. Integrating powers of sine and cosine.
Solution 1.
Solution 2. Video solution
Subsection 6.3.2 Integrals of the form \(\int\sin(mx)\sin(nx)\, dx\text{,}\) \(\int \cos(mx)\cos(nx)\, dx\text{,}\) and \(\int \sin(mx)\cos(nx)\, dx\)
Example 6.3.7. Integrating products of \(\sin(mx)\) and \(\cos(nx)\).
Solution 1.
Solution 2. Video solution
Subsection 6.3.3 Integrals of the form \(\int\tan^m(x) \sec^n(x) \, dx\)
- \(\frac{d}{dx}(\tan(x) ) = \sec^2(x)\text{,}\)
- \(\frac{d}{dx}(\sec(x) ) = \sec(x) \tan(x)\text{,}\)
- \(1+\tan^2(x) = \sec^2(x)\) (the Pythagorean Theorem).
Key Idea 6.3.8. Integrals Involving Powers of Tangent and Secant.
- If \(n\) is even, then \(n=2k\) for some integer \(k\text{.}\) Rewrite \(\sec^n(x)\) as\begin{align*} \sec^n(x) \amp = \sec^{2k}(x)\\ \amp = \sec^{2k-2}(x) \sec^2(x)\\ \amp = (1+\tan^2(x) )^{k-1}\sec^2(x)\text{.} \end{align*}Then\begin{align*} \int\tan^m(x) \sec^n(x) \, dx \amp =\int\tan^m(x) (1+\tan^2(x) )^{k-1}\sec^2(x) \, dx\\ \amp =\int u^m(1+u^2)^{k-1}\,du\text{,} \end{align*}where \(u = \tan(x)\) and \(du = \sec^2(x) \, dx\text{.}\)
- If \(m\) is odd, then \(m=2k+1\) for some integer \(k\text{.}\) Rewrite \(\tan^m(x) \sec^n(x)\) as\begin{align*} \tan^m(x) \sec^n(x) \amp = \tan^{2k+1}(x) \sec^n(x)\\ \amp = \tan^{2k}(x) \sec^{n-1}(x) \sec(x) \tan(x)\\ \amp = (\sec^2(x) -1)^k\sec^{n-1}(x) \sec(x) \tan(x)\text{.} \end{align*}Then\begin{align*} \int\tan^m(x) \sec^n(x) \, dx \amp =\int(\sec^2(x) -1)^k\sec^{n-1}(x) \sec(x) \tan(x) \, dx\\ \amp = \int(u^2-1)^ku^{n-1}\,du\text{,} \end{align*}where \(u = \sec(x)\) and \(du = \sec(x) \tan(x) \, dx\text{.}\)
- If \(n\) is odd and \(m\) is even, then \(m=2k\) for some integer \(k\text{.}\) Convert \(\tan^m(x)\) to \((\sec^2(x) -1)^k\text{.}\) Expand the new integrand and use Integration By Parts, with \(dv = \sec^2(x) \, dx\text{.}\)
- If \(m\) is even and \(n=0\text{,}\) rewrite \(\tan^m(x)\) as\begin{align*} \tan^m(x) \amp = \tan^{m-2}(x) \tan^2(x)\\ \amp = \tan^{m-2}(x) (\sec^2(x) -1)\\ \amp = \tan^{m-2}\sec^2(x) -\tan^{m-2}(x)\text{.} \end{align*}So\begin{equation*} \int\tan^m(x) \, dx = \underbrace{\int\tan^{m-2}\sec^2(x) \, dx}_{\text{ apply rule 1 } } - \underbrace{\int\tan^{m-2}(x) \, dx}_{\text{ apply rule 4 again } }\text{.} \end{equation*}
Example 6.3.9. Integrating powers of tangent and secant.
Solution 1.
Now substitute, with \(u=\tan(x)\text{,}\) with \(du = \sec^2(x) \, dx\text{.}\)
\begin{align*} \amp =\int u^2\big(1+u^2\big)^2\,du\\ \end{align*}We leave the integration and subsequent substitution to the reader. The final answer is
\begin{align*} \amp =\frac13\tan^3(x) +\frac25\tan^5(x) +\frac17\tan^7(x) +C\text{.} \end{align*}Solution 2. Video solution
Example 6.3.11. Integrating powers of tangent and secant.
Solution 1.
This new integral also requires applying Rule 3 of Key Idea 6.3.8:
\begin{align*} \int \sec^3(x)\, dx\amp = \sec(x) \tan(x) - \int \sec(x) \big(\sec^2(x) -1\big)\, dx\\ \amp = \sec(x) \tan(x) - \int \sec^3(x) \, dx + \int \sec(x) \, dx\\ \amp = \sec(x) \tan(x) -\int \sec^3(x) \, dx + \ln\abs{\sec(x) +\tan(x) }\\ \end{align*}In previous applications of Integration by Parts, we have seen where the original integral has reappeared in our work. We resolve this by adding \(\int \sec^3(x) \, dx\) to both sides, giving:
\begin{align*} 2\int \sec^3(x) \, dx \amp = \sec(x) \tan(x) + \ln\abs{\sec(x) +\tan(x) }\\ \int \sec^3(x) \, dx \amp = \frac12\Big(\sec(x) \tan(x) + \ln\abs{\sec(x) +\tan(x) }\Big)+C \end{align*}Solution 2. Video solution
Example 6.3.14. Integrating powers of tangent and secant.
Solution 1.
Integrate the first integral with substitution, \(u=\tan(x)\text{;}\) integrate the second by employing rule Rule 4 again.
\begin{align*} \amp = \frac15\tan^5(x) -\int\tan^2(x) \tan^2(x) \, dx\\ \amp = \frac15\tan^5(x) -\int\tan^2(x) \big(\sec^2(x) -1\big)\, dx\\ \amp = \frac15\tan^5(x) -\underbrace{\int\tan^2(x) \sec^2(x) \, dx}_{a} + \underbrace{\int\tan^2(x) \, dx}_b\\ \end{align*}Again, use substitution (\(u=\tan(x)\)) for the first integral (a) and Rule 4 for the second (b).
\begin{align*} \amp = \frac15\tan^5(x) -\frac13\tan^3(x) +\int\big(\sec^2(x) -1\big)\, dx\\ \int \tan^6(x)\, dx\amp = \frac15\tan^5(x) -\frac13\tan^3(x) +\tan(x) - x+C\text{.} \end{align*}Solution 2. Video solution
Exercises 6.3.4 Exercises
Terms and Concepts
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